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Rings: Eisenstein criterion

  1. May 26, 2012 #1
    My statement of Eisenstein's criterion is the following:

    Let [itex]R[/itex] be an integral domain, [itex]P[/itex] be prime ideal of [itex]R[/itex] and [itex]f(x) = a_0 + a_1x + ... + a_n x^n \in R[x][/itex].

    Suppose
    (1) [itex]a_0 , a_1 , ... , a_{n-1} \in P[/itex]
    (2) [itex]a_0 \in P[/itex] but [itex]a_0 \not\in P^2[/itex]
    (3) [itex]a_n \not\in P[/itex]

    Then [itex]f[/itex] has no divisors of degree [itex]d[/itex] such that [itex]1\leqslant d \leqslant n-1[/itex]. In particular if [itex]f[/itex] is primitive and (1)-(3) hold then [itex]f[/itex] is irreducible.


    I would like to see an example of how we can use this criterion in the Gaussian integers [itex]R= \mathbb{Z}[/itex].

    I know [itex]1+i[/itex] is a Gaussian prime so can I anyone give me a quick example of a polynomial with coefficients in [itex]\mathbb{Z}[/itex] and how to use this criterion?

    I know how to use it when [itex]R=\mathbb{Z}[/itex], for example to show [itex]f(x)= x^2 -1 \in \mathbb{Z}[x][/itex] is irreducible we just check that for a prime [itex]p[/itex]: [itex]p | a_n[/itex], [itex]p | a_i[/itex] for all [itex]i<n[/itex] and [itex]p^2 \not | a_0[/itex] . I'm confused as to how to use the version of it above for a polynomial in [itex]\mathbb{Z}[x][/itex].
     
    Last edited: May 26, 2012
  2. jcsd
  3. May 26, 2012 #2



    What about the polynomial [itex]\,x^2+(1+i)\in\left(\mathbb{Z}\right)[x]\,[/itex] ?

    BTW, in your example, [itex]\,x^2-1\in\mathbb{Z}[x]\,[/itex] is reducible...;>)

    DonAntonio
     
  4. May 27, 2012 #3


    Is the polynomial [tex]f(x) = x^7 + (3-i)x^2 + (3+4i)x + (4+2i) \in \mathbb{Z}[x][/tex] irreducible?

    [itex]2+i[/itex] is a Gauassian prime isn't it? And 2+i does not divide 1, 2+i | 3-i , 2+i | 3+4i , 2+i | 4+2i and (2+i)^2 = 3+4i which does not divide 4+2i.
     
    Last edited: May 27, 2012
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